The university skydiving club has asked you to plan a stunt for an air show. In this stunt, two skydivers will step out of opposite sides of a stationary hot air balloon 2,500m above the ground. the second skydiver will leave the ballon 20 seconds after the first skydiver but you want them both to land on the ground at the same time. the show is planned for a day with no wind so assume all motion is vertical. to ger rough idea of the situation, assume that a skydiver will fall with a constant acceleration of 9.8m/s^2 before the parachute opens. As soon as the parachute is opened, the skydiver falls with a constant velocity of 3.2m/s. if the first skydiver waits 3s after stepping out of the balloon before opening her parachute, How long must the second skydiver wait after leaving the balloon before opening his parachute?
I tried solving for time with the quadratic formula, but I am unsure of what procedure to take. Thanks for the help
To solve this problem, we can use the equations of motion for free fall. First, let's calculate the total time it takes for the first skydiver to reach the ground.
The first skydiver falls with a constant acceleration of 9.8 m/s^2 for 3 seconds before opening her parachute. During this time, her initial velocity is 0 m/s, and the distance she falls can be calculated using the equation:
d = (1/2) * a * t^2
where d is the distance fallen, a is the acceleration, and t is the time. Plugging in the values:
d = (1/2) * 9.8 * (3^2)
d = 44.1 m
After opening her parachute, the skydiver falls with a constant velocity of 3.2 m/s. To determine how long it takes for her to reach the ground from this point, we can use the equation:
d = v * t
where d is the distance fallen, v is the velocity, and t is the time. Plugging in the values:
2500 - 44.1 = 3.2 * t
t = (2500 - 44.1) / 3.2
t ≈ 775.94 seconds
Now we can calculate the time it takes for the second skydiver to reach the ground. The only difference is that this skydiver waits 20 seconds before jumping out of the balloon. Therefore, the time the second skydiver spends in free fall will be 20 seconds less. The equation becomes:
t = (2500 - 44.1) / 3.2 - 20
t ≈ 755.94 seconds
So, the second skydiver must wait approximately 755.94 seconds (or 12 minutes and 35.94 seconds) after leaving the balloon before opening his parachute in order to land at the same time as the first skydiver.
To solve this problem, we can use the kinematic equation for motion with constant acceleration:
d = vt + (1/2)at^2
where d is the distance traveled, v is the initial velocity, a is the acceleration, and t is the time.
Let's solve this step-by-step:
1. The first skydiver falls with a constant acceleration of 9.8 m/s^2 before opening her parachute. The initial velocity (v) is 0, since she starts from rest. The distance traveled (d) is the height of the balloon - 2,500 m.
d = vt + (1/2)at^2
2,500 = 0t + (1/2)(9.8)t^2
2. Simplify the equation:
2,500 = 4.9t^2
3. Divide both sides of the equation by 4.9:
510.2 = t^2
4. Take the square root of both sides:
t ≈ 22.61 seconds
So the first skydiver takes approximately 22.61 seconds to reach the ground.
5. The second skydiver waits 20 seconds after the first skydiver before leaving the balloon. So the time for the second skydiver is:
t2 = 22.61 + 20
t2 = 42.61 seconds
Therefore, the second skydiver must wait approximately 42.61 seconds after leaving the balloon before opening his parachute in order to land on the ground at the same time as the first skydiver.